# Group/Examples/inv x = 1 - x

## Theorem

Let $S = \set {x \in \R: 0 < x < 1}$.

Then an operation $\circ$ can be found such that $\struct {S, \circ}$ is a group such that the inverse of $x \in S$ is $1 - x$.

## Proof

Define $f: \openint 0 1 \to \R$ by:

- $\map f x := \map \ln {\dfrac {1 - x} x}$

Let us show that $f$ is a bijection by constructing an inverse mapping $g: \R \to \openint 0 1$:

- $\map g z := \dfrac 1 {1 + \exp z}$

### Lemma 1

- $\map {f \circ g} x = x$

### Lemma 2

- $\map {g \circ f} x = x$

This needs considerable tedious hard slog to complete it.In particular: Might also be necessary to demonstrate the domain of $f$ is the image set of $g$ and vice versa.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

Thus $f$ is a bijection.

Let $\struct {\R, +}$ be the additive group on $\R$.

Now define $\circ := +_f$ to be the operation induced on $\openint 0 1$ by $f$ and $+$:

- $x \circ y := \map {f^{-1} } {\map f x + \map f y}$

Let us determine the behaviour of $\circ$ more explicitly:

\(\ds x \circ y\) | \(=\) | \(\ds \map g {\map f x + \map f y}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac 1 {1 + \map \exp {\map \log {\frac {1 - x} x} + \map \log {\frac {1 - y} y} } }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac 1 {1 + \map \exp {\map \log {\frac {1 - x} x} } \map \exp {\map \log {\frac {1 - y} y} } }\) | Exponential of Sum | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac 1 {1 + \paren {\frac {1 - x} x} \paren {\frac {1 - y} y} }\) | Exponential of Natural Logarithm |

This article needs to be linked to other articles.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{MissingLinks}}` from the code. |

We see that $\circ$ is commutative.

Let $x = \dfrac 1 2$, $\dfrac {1 - x} x = 1$ so that:

\(\ds \dfrac 1 2 \circ y\) | \(=\) | \(\ds \dfrac 1 {1 + \paren {\frac {1 - y} y} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac 1 {\frac y y + \frac {1 - y} y}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac 1 {\frac 1 y}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds y\) |

so that $\dfrac 1 2$ is the identity element for $\circ$.

Furthermore, putting $y = 1 - x$, the following obtains:

\(\ds \frac 1 {1 + \paren {\frac {1 - x} x} \paren {\frac x {1 - x} } }\) | \(=\) | \(\ds \frac 1 {1 + 1}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac 1 2\) |

establishing $1 - x$ to be the inverse of $x$, as desired.

That $\circ$ in fact determines a group on $\openint 0 1$ follows from Pullback of Group is Group.

$\blacksquare$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 26 \mu$